Question

A sample containing the amino acid alanine plus inert matter is analyzed by the Kjeldahl methold. A 2.00-g sample is digested, the NH₃ is distilled and collected in 50.0 mL of 0.150 M H₂SO₄, and a volume of 9.0 mL of 0.100 M NaOH is required for back-titration. Calculate the percent alanine in the sample.

Answer 1

No of moles of H2SO4 = Molarity x volume in Litres = 0.15 M x 0.05 mol = 0.0075 mol

No of moles of NaOH = Molarity x volume in Litres = 0.1 M x 0.009 mol = 0.0009 mol

2NH3 + H2SO4 --------------> (NH4)2SO4

We know that

moles of H2SO4 reacted= moles of H2SO4 taken - ( moles of H2SO4 back titrated)

moles of H2SO4 reacted= moles of H2SO4 taken - ( moles of NaOH/2)--- (1)

moles of NH3/2 = moles of H2SO4 reacted ------ (2)

From (1) and (2),

moles of NH3/2 = moles of H2SO4 - (moles of NaOH/2)

                       = 0.0075 mol - (0.0009 mol/2)

                       = 0.00705 mol

Then,

moles of NH3 = 0.00705 mol x 2 = 0.0141 mol

But,

No of moles of NH3 = No of moles of Nitrogen = No of moles of alanine = 0.0141 mol

Hence,

mass of alanine = moles of alanine x molar mass of alanine

                         = 0.0141 mol x 89 g/mol

                         = 1.2549 g

---------------------------------------------------------------------------

Therefore,

percent alanine in the sample = (1.2549 g/ 2 g) x 100

                                           = 62.7 %